Trigonometric equations of increased complexity with solutions. Trigonometric equations

Lesson and presentation on the topic: "Solving simple trigonometric equations"

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What we will study:
1. What are trigonometric equations?

3. Two main methods for solving trigonometric equations.
4. Homogeneous trigonometric equations.
5. Examples.

What are trigonometric equations?

Guys, we have already studied arcsine, arccosine, arctangent and arccotangent. Now let's look at trigonometric equations in general.

Trigonometric equations– an equation in which a variable is contained under the sign of a trigonometric function.

Let us repeat the form of solving the simplest trigonometric equations:

1)If |a|≤ 1, then the equation cos(x) = a has a solution:

X= ± arccos(a) + 2πk

2) If |a|≤ 1, then sin equation(x) = a has a solution:

3) If |a| > 1, then the equation sin(x) = a and cos(x) = a have no solutions 4) The equation tg(x)=a has a solution: x=arctg(a)+ πk

5) The equation ctg(x)=a has a solution: x=arcctg(a)+ πk

For all formulas k is an integer

The simplest trigonometric equations have the form: T(kx+m)=a, T is some trigonometric function.

Example.

Solve the equations: a) sin(3x)= √3/2

Solution:

A) Let us denote 3x=t, then we will rewrite our equation in the form:

The solution to this equation will be: t=((-1)^n)arcsin(√3 /2)+ πn.

From the table of values we get: t=((-1)^n)×π/3+ πn.

Let's return to our variable: 3x =((-1)^n)×π/3+ πn,

Then x= ((-1)^n)×π/9+ πn/3

Answer: x= ((-1)^n)×π/9+ πn/3, where n is an integer. (-1)^n – minus one to the power of n.

More examples of trigonometric equations.

Solve the equations: a) cos(x/5)=1 b)tg(3x- π/3)= √3

Solution:

A) This time let’s move directly to calculating the roots of the equation right away:

X/5= ± arccos(1) + 2πk. Then x/5= πk => x=5πk

Answer: x=5πk, where k is an integer.

B) We write it in the form: 3x- π/3=arctg(√3)+ πk. We know that: arctan(√3)= π/3

3x- π/3= π/3+ πk => 3x=2π/3 + πk => x=2π/9 + πk/3

Answer: x=2π/9 + πk/3, where k is an integer.

Solve the equations: cos(4x)= √2/2. And find all the roots on the segment.

Solution:

We'll decide in general view our equation: 4x= ± arccos(√2/2) + 2πk

4x= ± π/4 + 2πk;

X= ± π/16+ πk/2;

Now let's see what roots fall on our segment. At k At k=0, x= π/16, we are in the given segment.
With k=1, x= π/16+ π/2=9π/16, we hit again.
For k=2, x= π/16+ π=17π/16, but here we didn’t hit, which means that for large k we also obviously won’t hit.

Answer: x= π/16, x= 9π/16

Two main solution methods.

We looked at the simplest trigonometric equations, but there are also more complex ones. To solve them, the method of introducing a new variable and the method of factorization are used. Let's look at examples.

Let's solve the equation:

Solution:
To solve our equation, we will use the method of introducing a new variable, denoting: t=tg(x).

As a result of the replacement we get: t 2 + 2t -1 = 0

Let's find the roots of the quadratic equation: t=-1 and t=1/3

Then tg(x)=-1 and tg(x)=1/3, we get the simplest trigonometric equation, let’s find its roots.

X=arctg(-1) +πk= -π/4+πk; x=arctg(1/3) + πk.

Answer: x= -π/4+πk; x=arctg(1/3) + πk.

An example of solving an equation

Solve equations: 2sin 2 (x) + 3 cos(x) = 0

Solution:

Let's use the identity: sin 2 (x) + cos 2 (x)=1

Our equation will take the form: 2-2cos 2 (x) + 3 cos (x) = 0

2 cos 2 (x) - 3 cos(x) -2 = 0

Let us introduce the replacement t=cos(x): 2t 2 -3t - 2 = 0

The solution to our quadratic equation is the roots: t=2 and t=-1/2

Then cos(x)=2 and cos(x)=-1/2.

Because cosine cannot take values greater than one, then cos(x)=2 has no roots.

For cos(x)=-1/2: x= ± arccos(-1/2) + 2πk; x= ±2π/3 + 2πk

Answer: x= ±2π/3 + 2πk

Homogeneous trigonometric equations.

Definition: Equations of the form a sin(x)+b cos(x) are called homogeneous trigonometric equations of the first degree.

Equations of the form

homogeneous trigonometric equations of the second degree.

To solve a homogeneous trigonometric equation of the first degree, divide it by cos(x): You cannot divide by the cosine if it is equal to zero, let's make sure that this is not the case:
Let cos(x)=0, then asin(x)+0=0 => sin(x)=0, but sine and cosine are not equal to zero at the same time, we get a contradiction, so we can safely divide by zero.

Solve the equation:
Example: cos 2 (x) + sin(x) cos(x) = 0

Solution:

Let's take out the common factor: cos(x)(c0s(x) + sin (x)) = 0

Then we need to solve two equations:

Cos(x)=0 and cos(x)+sin(x)=0

Cos(x)=0 at x= π/2 + πk;

Consider the equation cos(x)+sin(x)=0 Divide our equation by cos(x):

1+tg(x)=0 => tg(x)=-1 => x=arctg(-1) +πk= -π/4+πk

Answer: x= π/2 + πk and x= -π/4+πk

How to solve homogeneous trigonometric equations of the second degree?
Guys, always follow these rules!

1. See what the coefficient a is equal to, if a=0 then our equation will take the form cos(x)(bsin(x)+ccos(x)), an example of the solution of which is on the previous slide

2. If a≠0, then you need to divide both sides of the equation by the cosine squared, we get:

We change the variable t=tg(x) and get the equation:

Solve example No.:3

Solve the equation:
Solution:

Let's divide both sides of the equation by the cosine square:

We change the variable t=tg(x): t 2 + 2 t - 3 = 0

Let's find the roots of the quadratic equation: t=-3 and t=1

Then: tg(x)=-3 => x=arctg(-3) + πk=-arctg(3) + πk

Tg(x)=1 => x= π/4+ πk

Answer: x=-arctg(3) + πk and x= π/4+ πk

Solve example No.:4

Solve the equation:

Solution:
Let's transform our expression:

We can solve such equations: x= - π/4 + 2πk and x=5π/4 + 2πk

Answer: x= - π/4 + 2πk and x=5π/4 + 2πk

Solve example no.:5

Solve the equation:

Solution:
Let's transform our expression:

Let us introduce the replacement tg(2x)=t:2 2 - 5t + 2 = 0

The solution to our quadratic equation will be the roots: t=-2 and t=1/2

Then we get: tg(2x)=-2 and tg(2x)=1/2
2x=-arctg(2)+ πk => x=-arctg(2)/2 + πk/2

2x= arctg(1/2) + πk => x=arctg(1/2)/2+ πk/2

Answer: x=-arctg(2)/2 + πk/2 and x=arctg(1/2)/2+ πk/2

Problems for independent solution.

1) Solve the equation

A) sin(7x)= 1/2 b) cos(3x)= √3/2 c) cos(-x) = -1 d) tg(4x) = √3 d) ctg(0.5x) = -1.7

2) Solve the equations: sin(3x)= √3/2. And find all the roots on the segment [π/2; π].

3) Solve the equation: cot 2 (x) + 2 cot (x) + 1 =0

4) Solve the equation: 3 sin 2 (x) + √3sin (x) cos(x) = 0

5) Solve the equation: 3sin 2 (3x) + 10 sin(3x)cos(3x) + 3 cos 2 (3x) =0

6) Solve the equation: cos 2 (2x) -1 - cos(x) =√3/2 -sin 2 (2x)

Trigonometric equations are not an easy topic. They are too diverse.) For example, these:

sin 2 x + cos3x = ctg5x

sin(5x+π /4) = cot(2x-π /3)

sinx + cos2x + tg3x = ctg4x

Etc...

But these (and all other) trigonometric monsters have two common and obligatory features. First - you won’t believe it - there are trigonometric functions in the equations.) Second: all expressions with x are found within these same functions. And only there! If X appears somewhere outside, For example, sin2x + 3x = 3, this will already be an equation of mixed type. Such equations require an individual approach. We will not consider them here.

We will not solve evil equations in this lesson either.) Here we will deal with the simplest trigonometric equations. Why? Yes because the solution any trigonometric equations consists of two stages. At the first stage, the evil equation is reduced to a simple one through a variety of transformations. On the second, this simplest equation is solved. No other way.

So, if you have problems at the second stage, the first stage does not make much sense.)

What do elementary trigonometric equations look like?

sinx = a

cosx = a

tgx = a

ctgx = a

Here A stands for any number. Any.

By the way, inside a function there may not be a pure X, but some kind of expression, like:

cos(3x+π /3) = 1/2

etc. This complicates life, but does not affect the method of solving a trigonometric equation.

How to solve trigonometric equations?

Trigonometric equations can be solved in two ways. The first way: using logic and the trigonometric circle. We will look at this path here. The second way - using memory and formulas - will be discussed in the next lesson.

The first way is clear, reliable, and difficult to forget.) It is good for solving trigonometric equations, inequalities, and all sorts of tricky non-standard examples. Logic is stronger than memory!)

Solving equations using a trigonometric circle.

We include elementary logic and the ability to use the trigonometric circle. Don't you know how? However... You will have a hard time in trigonometry...) But it doesn’t matter. Take a look at the lessons "Trigonometric circle...... What is it?" and "Measuring angles on a trigonometric circle." Everything is simple there. Unlike textbooks...)

Oh, you know!? And even mastered “Practical work with the trigonometric circle”!? Congratulations. This topic will be close and understandable to you.) What is especially pleasing is that the trigonometric circle does not care what equation you solve. Sine, cosine, tangent, cotangent - everything is the same for him. There is only one solution principle.

So we take any elementary trigonometric equation. At least this:

cosx = 0.5

We need to find X. Speaking in human language, you need find the angle (x) whose cosine is 0.5.

How did we previously use the circle? We drew an angle on it. In degrees or radians. And right away saw trigonometric functions of this angle. Now let's do the opposite. Let's draw a cosine on the circle equal to 0.5 and immediately we'll see corner. All that remains is to write down the answer.) Yes, yes!

Draw a circle and mark the cosine equal to 0.5. On the cosine axis, of course. Like this:

Now let's draw the angle that this cosine gives us. Hover your mouse over the picture (or touch the picture on your tablet), and you'll see this very corner X.

The cosine of which angle is 0.5?

x = π /3

cos 60°= cos( π /3) = 0,5

Some people will chuckle skeptically, yes... Like, was it worth making a circle when everything is already clear... You can, of course, chuckle...) But the fact is that this is an erroneous answer. Or rather, insufficient. Circle connoisseurs understand that there are a whole bunch of other angles here that also give a cosine of 0.5.

If you turn the moving side OA full turn, point A will return to its original position. With the same cosine equal to 0.5. Those. the angle will change by 360° or 2π radians, and cosine - no. The new angle 60° + 360° = 420° will also be a solution to our equation, because

Such full revolutions you can wind up an infinite number... And all these new angles will be solutions to our trigonometric equation. And they all need to be written down somehow in response. All. Otherwise, the decision does not count, yes...)

Mathematics can do this simply and elegantly. Write down in one short answer infinite set decisions. Here's what it looks like for our equation:

x = π /3 + 2π n, n ∈ Z

I'll decipher it. Still write meaningfully It’s more pleasant than stupidly drawing some mysterious letters, right?)

π /3 - this is the same corner that we saw on the circle and determined according to the cosine table.

2π is one complete revolution in radians.

n - this is the number of complete ones, i.e. whole rpm It is clear that n can be equal to 0, ±1, ±2, ±3.... and so on. As indicated by the short entry:

n ∈ Z

n belongs ( ∈ ) set of integers ( Z ). By the way, instead of the letter n letters may well be used k, m, t etc.

This notation means you can take any integer n . At least -3, at least 0, at least +55. Whatever you want. If you substitute this number into the answer, you will get a specific angle, which will definitely be the solution to our harsh equation.)

Or, in other words, x = π /3 is the only root of an infinite set. To get all the other roots, it is enough to add any number of full revolutions to π /3 ( n ) in radians. Those. 2πn radian.

All? No. I deliberately prolong the pleasure. To remember better.) We received only part of the answers to our equation. I will write this first part of the solution like this:

x 1 = π /3 + 2π n, n ∈ Z

x 1 - not just one root, but a whole series of roots, written down in a short form.

But there are also angles that also give a cosine of 0.5!

Let's return to our picture from which we wrote down the answer. Here she is:

Hover your mouse over the image and we see another angle that also gives a cosine of 0.5. What do you think it is equal to? The triangles are the same... Yes! It is equal to the angle X , only delayed in the negative direction. This is the corner -X. But we have already calculated x. π /3 or 60°. Therefore, we can safely write:

x 2 = - π /3

Well, of course, we add all the angles that are obtained through full revolutions:

x 2 = - π /3 + 2π n, n ∈ Z

That's all now.) On the trigonometric circle we saw(who understands, of course)) All angles that give a cosine of 0.5. And we wrote down these angles in a short mathematical form. The answer resulted in two infinite series of roots:

x 1 = π /3 + 2π n, n ∈ Z

x 2 = - π /3 + 2π n, n ∈ Z

This is the correct answer.

Hope, general principle for solving trigonometric equations using a circle is clear. We mark the cosine (sine, tangent, cotangent) from the given equation on a circle, draw the angles corresponding to it and write down the answer. Of course, we need to figure out what corners we are saw on the circle. Sometimes it's not so obvious. Well, I said that logic is required here.)

For example, let's look at another trigonometric equation:

Please take into account that the number 0.5 is not the only possible number in equations!) It’s just more convenient for me to write it than roots and fractions.

We work according to the general principle. We draw a circle, mark (on the sine axis, of course!) 0.5. We draw all the angles corresponding to this sine at once. We get this picture:

Let's deal with the angle first X in the first quarter. We recall the table of sines and determine the value of this angle. It's a simple matter:

x = π /6

We remember about full revolutions and, with clear conscience, we write down the first series of answers:

x 1 = π /6 + 2π n, n ∈ Z

Half the job is done. But now we need to determine second corner... It's trickier than using cosines, yes... But logic will save us! How to determine the second angle through x? Yes Easy! The triangles in the picture are the same, and the red corner X equal to angle X . Only it is counted from the angle π in the negative direction. That’s why it’s red.) And for the answer we need an angle, measured correctly, from the positive semi-axis OX, i.e. from an angle of 0 degrees.

We hover the cursor over the drawing and see everything. I removed the first corner so as not to complicate the picture. The angle we are interested in (drawn in green) will be equal to:

π - x

X we know this π /6 . Therefore, the second angle will be:

π - π /6 = 5π /6

Again we remember about adding full revolutions and write down the second series of answers:

x 2 = 5π /6 + 2π n, n ∈ Z

That's all. A complete answer consists of two series of roots:

x 1 = π /6 + 2π n, n ∈ Z

x 2 = 5π /6 + 2π n, n ∈ Z

Tangent and cotangent equations can be easily solved using the same general principle for solving trigonometric equations. If, of course, you know how to draw tangent and cotangent on a trigonometric circle.

In the examples above, I used the table value of sine and cosine: 0.5. Those. one of those meanings that the student knows must. Now let's expand our capabilities to all other values. Decide, so decide!)

So, let's say we need to solve this trigonometric equation:

There is no such cosine value in the short tables. We coldly ignore this terrible fact. Draw a circle, mark 2/3 on the cosine axis and draw the corresponding angles. We get this picture.

Let's look, first, at the angle in the first quarter. If only we knew what x is equal to, we would immediately write down the answer! We don’t know... Failure!? Calm! Mathematics does not leave its own people in trouble! She came up with arc cosines for this case. Do not know? In vain. Find out, It's a lot easier than you think. There is not a single tricky spell about “inverse trigonometric functions” on this link... This is superfluous in this topic.

If you are in the know, just say to yourself: “X is an angle whose cosine is equal to 2/3.” And immediately, purely by the definition of arc cosine, we can write:

We remember about the additional revolutions and calmly write down the first series of roots of our trigonometric equation:

x 1 = arccos 2/3 + 2π n, n ∈ Z

The second series of roots for the second angle is almost automatically written down. Everything is the same, only X (arccos 2/3) will be with a minus:

x 2 = - arccos 2/3 + 2π n, n ∈ Z

And that's it! This is the correct answer. Even easier than with table values. There is no need to remember anything.) By the way, the most attentive will notice that this picture shows the solution through the arc cosine in essence, no different from the picture for the equation cosx = 0.5.

Exactly! General principle That's why it's common! I deliberately drew two almost identical pictures. The circle shows us the angle X by its cosine. Whether it is a tabular cosine or not is unknown to everyone. What kind of angle this is, π /3, or what arc cosine is - that’s up to us to decide.

Same song with sine. For example:

Draw a circle again, mark the sine equal to 1/3, draw the angles. This is the picture we get:

And again the picture is almost the same as for the equation sinx = 0.5. Again we start from the corner in the first quarter. What is X equal to if its sine is 1/3? No problem!

Now the first pack of roots is ready:

x 1 = arcsin 1/3 + 2π n, n ∈ Z

Let's deal with the second angle. In the example with a table value of 0.5, it was equal to:

π - x

It will be exactly the same here too! Only x is different, arcsin 1/3. So what!? You can safely write down the second pack of roots:

x 2 = π - arcsin 1/3 + 2π n, n ∈ Z

This is a completely correct answer. Although it doesn't look very familiar. But it’s clear, I hope.)

This is how trigonometric equations are solved using a circle. This path is clear and understandable. It is he who saves in trigonometric equations with the selection of roots on a given interval, in trigonometric inequalities - they are generally solved almost always in a circle. In short, in any tasks that are a little more difficult than standard ones.

Let's apply knowledge in practice?)

Solve trigonometric equations:

First, simpler, straight from this lesson.

Now it's more complicated.

Hint: here you will have to think about the circle. Personally.)

And now they are outwardly simple... They are also called special cases.

sinx = 0

sinx = 1

cosx = 0

cosx = -1

Hint: here you need to figure out in a circle where there are two series of answers and where there is one... And how to write one instead of two series of answers. Yes, so that not a single root from an infinite number is lost!)

Well, very simple):

sinx = 0,3

cosx = π

tgx = 1,2

ctgx = 3,7

Hint: here you need to know what arcsine and arccosine are? What is arctangent, arccotangent? The simplest definitions. But you don’t need to remember any table values!)

The answers are, of course, a mess):

x 1= arcsin0,3 + 2π n, n ∈ Z
x 2= π - arcsin0.3 + 2

Not everything works out? Happens. Read the lesson again. Only thoughtfully(there is such an outdated word...) And follow the links. The main links are about the circle. Without it, trigonometry is like crossing the road blindfolded. Sometimes it works.)

If you like this site...

By the way, I have a couple more interesting sites for you.)

You can practice solving examples and find out your level. Testing with instant verification. Let's learn - with interest!)

You can get acquainted with functions and derivatives.

Requires knowledge of the basic formulas of trigonometry - the sum of the squares of sine and cosine, the expression of tangent through sine and cosine, and others. For those who have forgotten them or do not know them, we recommend reading the article "".
So, we know the basic trigonometric formulas, it's time to use them in practice. Solving trigonometric equations with the right approach - quite exciting activity, like, for example, solving a Rubik's cube.

Based on the name itself, it is clear that a trigonometric equation is an equation in which the unknown is under the sign of the trigonometric function.
There are so-called simplest trigonometric equations. Here's what they look like: sinx = a, cos x = a, tan x = a. Let's consider how to solve such trigonometric equations, for clarity we will use the already familiar trigonometric circle.

sinx = a

cos x = a

tan x = a

cot x = a

Any trigonometric equation is solved in two stages: we reduce the equation to its simplest form and then solve it as a simple trigonometric equation.
There are 7 main methods by which trigonometric equations are solved.

Variable substitution and substitution method

Solve the equation 2cos 2 (x + /6) – 3sin( /3 – x) +1 = 0

Using the reduction formulas we get:

2cos 2 (x + /6) – 3cos(x + /6) +1 = 0

Replace cos(x + /6) with y to simplify and get the usual quadratic equation:

2y 2 – 3y + 1 + 0

The roots of which are y 1 = 1, y 2 = 1/2

Now let's go in reverse order

We substitute the found values of y and get two answer options:

Solving trigonometric equations through factorization

How to solve the equation sin x + cos x = 1?

Let's move everything to the left so that 0 remains on the right:

sin x + cos x – 1 = 0

Let us use the identities discussed above to simplify the equation:

sin x - 2 sin 2 (x/2) = 0

Let's factorize:

2sin(x/2) * cos(x/2) - 2 sin 2 (x/2) = 0

2sin(x/2) * = 0

We get two equations

Reduction to a homogeneous equation

An equation is homogeneous with respect to sine and cosine if all its terms are relative to the sine and cosine of the same degree of the same angle. To solve a homogeneous equation, proceed as follows:

a) transfer all its members to the left side;

b) take all common factors out of brackets;

c) equate all factors and brackets to 0;

d) a homogeneous equation of a lower degree is obtained in brackets, which in turn is divided into a sine or cosine of a higher degree;

e) solve the resulting equation for tg.

Solve the equation 3sin 2 x + 4 sin x cos x + 5 cos 2 x = 2

Let's use the formula sin 2 x + cos 2 x = 1 and get rid of the open two on the right:

3sin 2 x + 4 sin x cos x + 5 cos x = 2sin 2 x + 2cos 2 x

sin 2 x + 4 sin x cos x + 3 cos 2 x = 0

Divide by cos x:

tg 2 x + 4 tg x + 3 = 0

Replace tan x with y and get a quadratic equation:

y 2 + 4y +3 = 0, whose roots are y 1 =1, y 2 = 3

From here we find two solutions to the original equation:

x 2 = arctan 3 + k

Solving equations through the transition to a half angle

Solve the equation 3sin x – 5cos x = 7

Let's move on to x/2:

6sin(x/2) * cos(x/2) – 5cos 2 (x/2) + 5sin 2 (x/2) = 7sin 2 (x/2) + 7cos 2 (x/2)

Let's move everything to the left:

2sin 2 (x/2) – 6sin(x/2) * cos(x/2) + 12cos 2 (x/2) = 0

Divide by cos(x/2):

tg 2 (x/2) – 3tg(x/2) + 6 = 0

Introduction of auxiliary angle

For consideration, let’s take an equation of the form: a sin x + b cos x = c,

where a, b, c are some arbitrary coefficients, and x is an unknown.

Let's divide both sides of the equation by:

Now the coefficients of the equation, according to trigonometric formulas, have the properties sin and cos, namely: their modulus is not more than 1 and the sum of squares = 1. Let us denote them respectively as cos and sin, where - this is the so-called auxiliary angle. Then the equation will take the form:

cos * sin x + sin * cos x = C

or sin(x + ) = C

The solution to this simplest trigonometric equation is

x = (-1) k * arcsin C - + k, where

It should be noted that the notations cos and sin are interchangeable.

Solve the equation sin 3x – cos 3x = 1

The coefficients in this equation are:

a = , b = -1, so divide both sides by = 2

Concept of solving trigonometric equations.

To solve a trigonometric equation, convert it into one or more basic trigonometric equations. Solving a trigonometric equation ultimately comes down to solving the four basic trigonometric equations.

Solving basic trigonometric equations.

There are 4 types of basic trigonometric equations:
sin x = a; cos x = a
tan x = a; ctg x = a
Solving basic trigonometric equations involves looking at different x positions on the unit circle, as well as using a conversion table (or calculator).
Example 1. sin x = 0.866. Using a conversion table (or calculator) you will get the answer: x = π/3. The unit circle gives another answer: 2π/3. Remember: all trigonometric functions are periodic, meaning their values repeat. For example, the periodicity of sin x and cos x is 2πn, and the periodicity of tg x and ctg x is πn. Therefore the answer is written as follows:
x1 = π/3 + 2πn; x2 = 2π/3 + 2πn.
Example 2. cos x = -1/2. Using a conversion table (or calculator) you will get the answer: x = 2π/3. The unit circle gives another answer: -2π/3.
x1 = 2π/3 + 2π; x2 = -2π/3 + 2π.
Example 3. tg (x - π/4) = 0.
Answer: x = π/4 + πn.
Example 4. ctg 2x = 1.732.
Answer: x = π/12 + πn.

Transformations used in solving trigonometric equations.

To transform trigonometric equations, algebraic transformations (factorization, reduction of homogeneous terms, etc.) and trigonometric identities are used.
Example 5: Using trigonometric identities, the equation sin x + sin 2x + sin 3x = 0 is converted to the equation 4cos x*sin (3x/2)*cos (x/2) = 0. Thus, the following basic trigonometric equations need to be solved: cos x = 0; sin(3x/2) = 0; cos(x/2) = 0.
Finding angles by known values functions.
- Before learning how to solve trigonometric equations, you need to learn how to find angles using known function values. This can be done using a conversion table or calculator.
- Example: cos x = 0.732. The calculator will give the answer x = 42.95 degrees. The unit circle will give additional angles, the cosine of which is also 0.732.
Set aside the solution on the unit circle.
- You can plot solutions to a trigonometric equation on the unit circle. Solutions to a trigonometric equation on the unit circle are the vertices of a regular polygon.
- Example: The solutions x = π/3 + πn/2 on the unit circle represent the vertices of the square.
- Example: The solutions x = π/4 + πn/3 on the unit circle represent the vertices of a regular hexagon.
Methods for solving trigonometric equations.
- If a given trigonometric equation contains only one trigonometric function, solve that equation as a basic trigonometric equation. If a given equation includes two or more trigonometric functions, then there are 2 methods for solving such an equation (depending on the possibility of its transformation).
  - Method 1.
- Transform this equation into an equation of the form: f(x)*g(x)*h(x) = 0, where f(x), g(x), h(x) are the basic trigonometric equations.
- Example 6. 2cos x + sin 2x = 0. (0< x < 2π)
- Solution. Using the double angle formula sin 2x = 2*sin x*cos x, replace sin 2x.
- 2cos x + 2*sin x*cos x = 2cos x*(sin x + 1) = 0. Now solve the two basic trigonometric equations: cos x = 0 and (sin x + 1) = 0.
- Example 7. cos x + cos 2x + cos 3x = 0. (0< x < 2π)
- Solution: Using trigonometric identities, transform this equation into an equation of the form: cos 2x(2cos x + 1) = 0. Now solve the two basic trigonometric equations: cos 2x = 0 and (2cos x + 1) = 0.
- Example 8. sin x - sin 3x = cos 2x. (0< x < 2π)
- Solution: Using trigonometric identities, transform this equation into an equation of the form: -cos 2x*(2sin x + 1) = 0. Now solve the two basic trigonometric equations: cos 2x = 0 and (2sin x + 1) = 0.
  - Method 2.
- Convert the given trigonometric equation into an equation containing only one trigonometric function. Then replace this trigonometric function with some unknown one, for example, t (sin x = t; cos x = t; cos 2x = t, tan x = t; tg (x/2) = t, etc.).
- Example 9. 3sin^2 x - 2cos^2 x = 4sin x + 7 (0< x < 2π).
- Solution. In this equation, replace (cos^2 x) with (1 - sin^2 x) (according to the identity). The transformed equation is:
- 3sin^2 x - 2 + 2sin^2 x - 4sin x - 7 = 0. Replace sin x with t. Now the equation looks like: 5t^2 - 4t - 9 = 0. This is a quadratic equation that has two roots: t1 = -1 and t2 = 9/5. The second root t2 does not satisfy the function range (-1< sin x < 1). Теперь решите: t = sin х = -1; х = 3π/2.
- Example 10. tg x + 2 tg^2 x = ctg x + 2
- Solution. Replace tg x with t. Rewrite the original equation as follows: (2t + 1)(t^2 - 1) = 0. Now find t and then find x for t = tan x.

Solving simple trigonometric equations.

Solving trigonometric equations of any level of complexity ultimately comes down to solving the simplest trigonometric equations. And in this best helper again it turns out to be a trigonometric circle.

Let's recall the definitions of cosine and sine.

The cosine of an angle is the abscissa (that is, the coordinate along the axis) of a point on the unit circle corresponding to a rotation through a given angle.

The sine of an angle is the ordinate (that is, the coordinate along the axis) of a point on the unit circle corresponding to a rotation through a given angle.

The positive direction of movement on the trigonometric circle is counterclockwise. A rotation of 0 degrees (or 0 radians) corresponds to a point with coordinates (1;0)

We use these definitions to solve simple trigonometric equations.

1. Solve the equation

This equation is satisfied by all values of the rotation angle that correspond to points on the circle whose ordinate is equal to .

Let's mark a point with ordinate on the ordinate axis:

Draw a horizontal line parallel to the x-axis until it intersects with the circle. We get two points lying on the circle and having an ordinate. These points correspond to rotation angles in and radians:

If we, leaving the point corresponding to the angle of rotation per radian, go around a full circle, then we will arrive at a point corresponding to the angle of rotation per radian and having the same ordinate. That is, this rotation angle also satisfies our equation. We can make as many “idle” revolutions as we like, returning to the same point, and all these angle values will satisfy our equation. The number of “idle” revolutions will be denoted by the letter (or). Since we can make these revolutions in both positive and negative directions, (or) can take on any integer values.

That is, the first series of solutions to the original equation has the form:

, , - set of integers (1)

Similarly, the second series of solutions has the form:

, Where , . (2)

As you might have guessed, this series of solutions is based on the point on the circle corresponding to the angle of rotation by .

These two series of solutions can be combined into one entry:

If we take (that is, even) in this entry, then we will get the first series of solutions.

If we take (that is, odd) in this entry, then we get the second series of solutions.

2. Now let's solve the equation

Since this is the abscissa of a point on the unit circle obtained by rotating through an angle, we mark the point with the abscissa on the axis:

Draw a vertical line parallel to the axis until it intersects with the circle. We will get two points lying on the circle and having an abscissa. These points correspond to rotation angles in and radians. Recall that when moving clockwise we get a negative rotation angle:

Let us write down two series of solutions:

(We get to the desired point by going from the main full circle, that is.

Let's combine these two series into one entry:

3. Solve the equation

The tangent line passes through the point with coordinates (1,0) of the unit circle parallel to the OY axis

Let's mark a point on it with an ordinate equal to 1 (we are looking for the tangent of which angles is equal to 1):

Let's connect this point to the origin of coordinates with a straight line and mark the points of intersection of the line with the unit circle. The intersection points of the straight line and the circle correspond to the angles of rotation on and :

Since the points corresponding to the rotation angles that satisfy our equation lie at a distance of radians from each other, we can write the solution this way:

4. Solve the equation

The line of cotangents passes through the point with the coordinates of the unit circle parallel to the axis.

Let's mark a point with abscissa -1 on the line of cotangents:

Let's connect this point to the origin of the straight line and continue it until it intersects with the circle. This straight line will intersect the circle at points corresponding to the angles of rotation in and radians: